# Math Help - Derivative help

1. ## Derivative help

Question:

f(x) = ln(cosx)

I need to find the derivative.

I'm unsure where to start. I tried the chain rule, and I get stuck, as follows:

$\frac {1}{x} (cosx)^0 (-sinx)$

$\frac {-sinx}{x}$

The answer is supposed to be -tanx

2. Originally Posted by Glitch
f(x) = ln(cosx)

I need to find the derivative.
make $u = \cos x \implies y = \ln u$

$\frac{du}{dx} = -\sin x$ and $\frac{dy}{du}= \frac{1}{u}$

$
\frac{dy}{dx}= \frac{dy}{du}\times \frac{du}{dx}= \frac{1}{u}\times -\sin x = \frac{-\sin x}{\cos x} = -\tan x
$

3. Interesting. So that's the chain rule? Your working out isn't very familiar to me.

4. Originally Posted by Glitch
Interesting. So that's the chain rule? Your working out isn't very familiar to me.
hi
if u have two functions of $x$, $f$ and $g$,how do you find the derivative of $f(g(x))$ ?

5. I find the derivative of g(x), then the derivative of f(g(x)), then multiply them. I think the bit that confuses me is since the function is the natural logarithm, you don't lower the power like I did in the OP. Or maybe you do, and I'm going nuts.

6. Originally Posted by Glitch
I find the derivative of g(x), then the derivative of f(g(x)), then multiply them. I think the bit that confuses me is since the function is the natural logarithm, you don't lower the power like I did in the OP. Or maybe you do, and I'm going nuts.
well ..
$\left (\ln f(x) \right )'=\frac{f'(x)}{f(x)}$
$\left (\ln \cos(x) \right )'=\frac{-\sin(x)}{\cos(x)}$
$(\cos(x))'=-\sin(x)$

7. Yeah, I made an error. Thanks for the help!

8. A general formula for the 'logarithmic derivative' of a function is...

d/dx ln f(x) = f'(x)/f(x) (1)

... and here is f(x) = cos x , so that...

Kind regards

$\chi$ $\sigma$

9. Originally Posted by Glitch
I find the derivative of g(x), then the derivative of f(g(x)), then multiply them. I think the bit that confuses me is since the function is the natural logarithm, you don't lower the power like I did in the OP. Or maybe you do, and I'm going nuts.
Why were you lowering the power, though? The derivative of ln{x} is 1/{x} and the derivative of cosx is -sinx. That should be sufficient:

$\dfrac{d}{dx}\left\{\ln(\cos{x})\right\} = \left(\dfrac{1}{\cos{x}}\right)(-\sin{x}) = -\dfrac{\sin{x}}{\cos{x}} = -\tan{x}.$